## Intermediate Algebra (6th Edition)

$x=\dfrac{\log\dfrac{1}{2}+\log5}{\log5}\\\\ x\approx0.5693$
Getting the logarithm of both sides and then using the properties of logarithms, the value of $x$ in the expression $5^{x-1}=\dfrac{1}{2}$ is \begin{array}{l} \log5^{x-1}=\log\dfrac{1}{2}\\\\ (x-1)\log5=\log\dfrac{1}{2}\\\\ x\log5-\log5=\log\dfrac{1}{2}\\\\ x\log5=\log\dfrac{1}{2}+\log5\\\\ x=\dfrac{\log\dfrac{1}{2}+\log5}{\log5}\\\\ x\approx0.5693 .\end{array}